# Find the angles between the vectors A with arrow and B with arrow given the following.

A with arrow = 2i - 5j, B with arrow = -5i + 1j

A with arrow = 4i + 6j, B with arrow = 1i - 5j

A with arrow = 7i + 3j, B with arrow = 3i - 7j

**Solution:**

(i) Given: the vectors** **A = 2i - 5j; B = -5i + 1j

From the dot product of vectors, We know that cos θ = (u.v)/|u||v|, Here, θ is the angle between vectors.

(u.v) = (2i - 5j ).(-5i + 1j) = (2)(-5) +(-5)(1) = -10 - 5 = -15

|u| = √(4 + 25) = √29

|v| = √(25 + 1)= √26

Cosθ = -15/(√29 × √26) = -0.546

θ = cos^{-1}(-0.549)

θ = 123°

(ii)** **u = 4i + 6j; v = 1i - 5j

We know that cosθ = (u.v)/|u||v|

(u.v) = ( 4i + 6j ).(i - 5j )

= (4)(1) + (6)(-5)

= 4 - 30

(u.v) = -26

|u| = √(16 + 36) = √52

|v| = √(1 + 25) = √26

Cosθ = -26/(√52 × √26) = 36.76

θ = cos^{-1}(36.76)

θ = 134.99°

(iii)** **u = 7i + 3j; v = 3i - 7j

We know that cosθ = (u.v)/|u||v|

(u.v) = (7i + 3j ).(3i - 7j)

= (7)(3) +(3)(-7)

= 21 - 21

(u.v) = 0

|u| = √(49 + 9) = √58

|v| = √(9 + 49) = √58

Cosθ = 0/(√58 × √58) = 0

θ = cos^{-1}(0)

θ = 90°

## Find the angles between the vectors A with arrow and B with arrow given the following.

A with arrow = 2i - 5j, B with arrow = -5i + 1j

A with arrow = 4i + 6j, B with arrow = 1i - 5j

A with arrow = 7i + 3j, B with arrow = 3i - 7j

**Summary:**

(i) The angle for A with arrow = 2i - 5j, B with arrow = -5i + 1j is 123°.

(ii) The angle for A with arrow = 4i + 6j, B with arrow = 1i - 5j is 135°.

(iii) The angle for A with arrow = 7i + 3j, B with arrow = 3i - 7jis 90°.

visual curriculum